Limit of a product of cosines minus 1, over x²
Evaluate the limit, in terms of the positive integer $n$: $$ L(n) = \lim_{x \to 0} \frac{\cos x\cos 2x\cdots\cos(nx) - 1}{x^2}. $$
Explanation of Method
Direct substitution gives $\frac{1-1}{0}=\frac00$. Since the numerator is a product of $n$ cosines minus $1$, taking the logarithm of the product turns it into a sum, where each term $\ln(\cos(kx))$ can be expanded using the small-angle approximation $1-\cos(kx)\sim\frac{(kx)^2}2$. Summing these contributions brings in the classic formula for the sum of the first $n$ squares, $\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$, which becomes the key ingredient of the final answer.
- Approximate each factor $\cos(kx)$ using its Taylor expansion for small $x$.
- Take the logarithm of the product, converting it to a sum, and use $\ln(1-y)\approx-y$.
- Sum the resulting series using the formula for $\sum_{k=1}^n k^2$.
- Exponentiate back and simplify $\dfrac{\text{product}-1}{x^2}$.
Step 1 · Approximate each cosine factor
For each $k=1,\dots,n$, the Taylor expansion of $\cos(kx)$ near $x=0$ is:
$$ \cos(kx) = 1-\frac{k^2x^2}{2}+O(x^4). $$
Step 2 · Take the logarithm of the product
Let $P(x) = \prod_{k=1}^n\cos(kx)$. Taking logarithms turns the product into a sum:
$$ \ln P(x) = \sum_{k=1}^n \ln\left(\cos(kx)\right) = \sum_{k=1}^n \ln\left(1-\frac{k^2x^2}2+O(x^4)\right). $$
Using $\ln(1-y)\approx-y$ for small $y$, with $y=\frac{k^2x^2}2+O(x^4)$:
$$ \ln P(x) \approx -\sum_{k=1}^n \frac{k^2x^2}{2} = -\frac{x^2}{2}\sum_{k=1}^n k^2. $$
Step 3 · Sum using the formula for $\sum k^2$
Using the standard identity $\displaystyle\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$:
$$ \ln P(x) \approx -\frac{x^2}{2}\cdot\frac{n(n+1)(2n+1)}{6} = -\frac{n(n+1)(2n+1)}{12}x^2. $$
Step 4 · Exponentiate and simplify
Exponentiating, and using $e^z\approx1+z$ for small $z$:
$$ P(x) = e^{\ln P(x)} \approx 1 - \frac{n(n+1)(2n+1)}{12}x^2. $$
So:
$$ P(x) - 1 \approx -\frac{n(n+1)(2n+1)}{12}x^2, $$
and dividing by $x^2$:
$$ L(n) = \lim_{x\to0}\frac{P(x)-1}{x^2} = -\frac{n(n+1)(2n+1)}{12}. $$
Summary Table
| Quantity | Expression |
|---|---|
| $\ln P(x)$ | $-\dfrac{x^2}{2}\displaystyle\sum_{k=1}^nk^2 = -\dfrac{n(n+1)(2n+1)}{12}x^2$ |
| $P(x)-1$ | $\approx-\dfrac{n(n+1)(2n+1)}{12}x^2$ |
| $L(n)$ | $-\dfrac{n(n+1)(2n+1)}{12}$ |
Additional Notes
- As a quick sanity check, $n=1$ gives $L(1)=-\frac{1\cdot2\cdot3}{12}=-\frac12$, matching the well-known single-cosine limit $\lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac12$ (equivalent to $1-\cos x\sim\frac{x^2}2$ seen elsewhere on this site).
- The technique of taking logarithms to turn a product into a sum is essential here: expanding the product $\prod\cos(kx)$ directly and collecting the $x^2$ terms would require the same computation, but the log-sum route makes the appearance of $\sum k^2$ transparent.
- This generalizes to products of any family of functions each behaving like $1-c_kx^2+O(x^4)$: the resulting limit of $\frac{\text{product}-1}{x^2}$ is always $-\sum_k c_k$, regardless of the specific functions involved.
Result
The value of the limit is: